B Quantum Entanglement information transmission idea to knock down....

[LozD]

TL;DR

Many entangled particle pairs are prepared with property A (1 or 0) at 99% probability of 1, and property B (1 or 0) at 1% probability of 1. The pairs are separated with 2 observers, X and Y, each having a particle of each pair. They pre-arrange that at a certain time: Y will "send a 0" by NOT measuring anything on his particles or Y will "send a 1" by measuring property B on every one of his particles. X will measure property A on her particles: If 99% = 1, Y sent a 1; if 50%, Y sent a 0.

I work in IT and am a layman in the quantum world. I have obviously misunderstood something in my amateur reading of quantum, but if someone could explain my mistake in the above scenario it might be very insightful for me! Forgive me if the terminology is not correct - or if indeed lay folks' enquiries on this forum are inappropriate then please ignore me!

Here's the above again with a bit more detail and the concepts that lead me to assume the main steps ...

1. I take a pair of particles with two properties (spin, velocity, whatever) A and B with values 1 or 0 and prepare them (entangle them): 99% A is 1, 1% B is 1.
2. Send one of the particles with another observer to another place.
3. Tell them that at a pre-arranged time they must either (a) not observe their particle if they wish to send a 0, or, (b) observe (measure) the B property if they wish to send a 1.
4. At the pre-arranged time (or just after!) I then observe my particle's A value.
5. If the sender *didn't* observe her particle then the A property of my particle will be 1 with the original prepared 99% probability, because I am the first observer of any property of the entangled pair. (Btw, thereafter, property B, if examined by me or the other observer, would only have a 50% chance, not 1%, of being 1, due to "complementarity" and the Heisenberg Uncertainty Principle, but this is not directly relevant here). So, per the pre-arranged plan, the remote observer has "sent a 0" if the probability of property A being 1 is 99%.
6. If the sender *did* observe her particle's B property then the A property of my particle will be 1 with only a 50% probability, because the sender has observed property B and due to "complementarity" and the Heisenberg's Uncertainty Principle the probability of property A being 1 has collapsed to only 50%.
7. By using multiple pairs of such entangled particles the probability difference of 99% vs 50% for 0 or 1 respectively will allow me to derive a 0 was sent if the number of 1's on A of my particles is near 99%, or a 1 was sent if the number of 1's is near 50%, with good accuracy.
   

This appears not to require any comparison value of A to be sent from the remote observer back to me after the above events which would necessarily happen at equal to or less than the speed of light via another communication method. Rather, the information arrives instantaneously - so long as the synchronous nature of the pre-arrangement protocol is reliable.

My reasoning around the "complementarity" and Heisenberg Uncertainty Principle pieces came from the narrative in the section entitled "Bohm's variant" in this link ...

https://en.m.wikipedia.org/wiki/EPR_paradox

Whatever axis their spins are measured along, they are always found to be opposite. In quantum mechanics, the x-spin and z-spin are "incompatible observables", meaning the Heisenberg uncertainty principle applies to alternating measurements of them: a quantum state cannot possesses a definite value for both of these variables. Suppose Alice measures the z-spin and obtains +z, so that the quantum state collapses into state I. Now, instead of measuring the z-spin as well, Bob measures the x-spin. According to quantum mechanics, when the system is in state I, Bob's x-spin measurement will have a 50% probability of producing +x and a 50% probability of -x. It is impossible to predict which outcome will appear until Bob actually performs the measurement. Therefore, Bob's positron will have a definite spin when measured along the same axis as Alice's electron, but when measured in the perpendicular axis its spin will be uniformly random. It seems as if information has propagated (faster than light) from Alice's apparatus to make Bob's positron assume a definite spin in the appropriate axis.

... and from the narrative in this excellent layman's explanation of Quantum Entanglement ...

https://www.quantamagazine.org/entanglement-made-simple-20160428/

Upon deeper reflection, the paradox dissolves further. Indeed, let us consider again the state of the second system, given that the first has been measured to be red. If we choose to measure the second q-on’s color, we will surely get red. But as we discussed earlier, when introducing complementarity, if we choose to measure a q-on’s shape, when it is in the “red” state, we will have equal probability to find a square or a circle. Thus, far from introducing a paradox, the EPR outcome is logically forced. It is, in essence, simply a repackaging of complementarity.

If anyone who knows what they're talking about has got a spare moment to put me straight, please fire away!

 

[LozD]

Just in case it's not clear what my issue is: as far as I have read, using Quantum Entanglement to communicate information, i.e. like data, e.g. 1's and 0's, is not possible with QE without the transmission of a comparison copy of the information making its way from sender to recipient via another non-QE route in parallel, that necessarily, due to Special Relativity, transmits at equal to or less than the speed of light. My above thought experiment scenario purports to contradict that limitation and does allow information transmission over the purely QE system instantaneously with no parallel, sub- speed-of-light, back-channel required.

 

[PeroK]

Summary:: Many entangled particle pairs are prepared with property A (1 or 0) at 99% probability of 1, and property B (1 or 0) at 1% probability of 1. The pairs are separated with 2 observers, X and Y, each having a particle of each pair. They pre-arrange that at a certain time: Y will "send a 0" by NOT measuring anything on his particles or Y will "send a 1" by measuring property B on every one of his particles. X will measure property A on her particles: If 99% = 1, Y sent a 1; if 50%, Y sent a 0.

The distribution of the results of X's measurements is independent of what Y does. There's no way for X to know whether Y undertook any measurements or not.

So, yes, you've fundamentally misundersood quantum entanglement.

 

[LozD]

OK, I thought entanglement meant the two particles share the same state? So if A measures the state of his particle whatever it is will be the state of B's particle, even though they won't know they've measured it without transmitting that by other means. So A could say to himself on measuring the state of his particle, "Aha, it's a 1. I've no idea if B has read it yet, but if she does, it will be a 1 also". However, as soon as that "first read" is done, all bets are off in terms of the other prepared states (velocity, spin, etc.) on the particle. They collapse to being 50% (if the possible values are binary for example) and are no longer the original probabilities when the particles were prepared and both observers informed of those probabilities.

 

[PeroK]

OK, I thought entanglement meant the two particles share the same state? So if A measures the state of his particle whatever it is will be the state of B's particle, even though they won't know they've measured it without transmitting that by other means. So A could say to himself on measuring the state of his particle, "Aha, it's a 1. I've no idea if B has read it yet, but if she does, it will be a 1 also".

Yes, but that's a not a message from A to B (or from B to A) as A and B have no control over the results of their own or the other's measurements.

 

[LozD]

Quite so, it's not a message. It's not on that B "channel" that the message is really being sent. The B state is merely there for Y to "measure or not at midnight" (or whatever time) and make the first "read" of this connected system. It doesn't matter what value is discovered on B by Y, it's irrelevant. And X will never even look at B either, doesn't need to. Instead X will look at A at 1 minute past midnight, which Y never touched at her end. If X discovers A is 99% chance 1, then X will know that Y didn't measure anything at midnight, specifically, Y didn't measure B. If X discovers A is 50% chance 1, then X will know that Y did measure B at midnight. A 1 or 0 has passed from X to Y through their pre-arrangement of midnight and 1 minute past midnight synchronous protocol.

 

[Nugatory]

OK, I thought entanglement meant the two particles share the same state?

Yes, they do have the same state. However, if we prepare the particles in the state such that A has a 99% chance of getting a 1 and a 1% chance of getting a zero and (for the sake of argument the entanglement is such that) B’s measurement on the same axis will have the same probabilities, then any measurement on any axis performed by B will be consistent with that state - no matter what A has done or will do.

As an aside, although the mathematical formalism allows me to write down and reason about that state, I know of no process that could produce it in the real world. Real experiments are done with the physically realizable state (the “singlet state” when we’re working with spin) in which A and B both have a 50% chance of getting either result.

 

[PeroK]

Quite so, it's not a message. It's not on that B "channel" that the message is really being sent. The B state is merely there for Y to "measure or not at midnight" (or whatever time) and make the first "read" of this connected system. It doesn't matter what value is discovered on B by Y, it's irrelevant. And X will never even look at B either, doesn't need to. Instead X will look at A at 1 minute past midnight, which Y never touched at her end. If X discovers A is 99% chance 1, then X will know that Y didn't measure anything at midnight, specifically, Y didn't measure B.

I've already said that is a fundamental misunderstanding. X will get A 99% of the time regardless of what Y does or does not do.

 

[DrChinese]

1. I take a pair of particles with two properties (spin, velocity, whatever) A and B with values 1 or 0 and prepare them (entangle them): 99% A is 1, 1% B is 1.

This is not an entangled state. In an entangled state, the particles' entangled properties do not have a specific value until they are measured. So you would actually have something more like:

Before measurement:
50% (A=1 + B=0) + 50% (A=0 + B=1)

After measurement:
100% (A=1, B=0) [or vice versa 100% (A=0, B=1)]

Note that this assumes you are measuring the 2 particles in the same manner. The order of measurement is not important to the statistical results.

 

[LozD]

According to the following paragraphs from the links I gave, they appear to contradict you - see my bold ...

Wiki link: -

Whatever axis their spins are measured along, they are always found to be opposite. In quantum mechanics, the x-spin and z-spin are "incompatible observables", meaning the Heisenberg uncertainty principle applies to alternating measurements of them: a quantum state cannot possesses a definite value for both of these variables. Suppose Alice measures the z-spin and obtains +z, so that the quantum state collapses into state I. Now, instead of measuring the z-spin as well, Bob measures the x-spin. According to quantum mechanics, when the system is in state I, Bob's x-spin measurement will have a 50% probability of producing +x and a 50% probability of -x. It is impossible to predict which outcome will appear until Bob actually performs the measurement. Therefore, Bob's positron will have a definite spin when measured along the same axis as Alice's electron, but when measured in the perpendicular axis its spin will be uniformly random. It seems as if information has propagated (faster than light) from Alice's apparatus to make Bob's positron assume a definite spin in the appropriate axis.

Quanta Magazine link: -

Upon deeper reflection, the paradox dissolves further. Indeed, let us consider again the state of the second system, given that the first has been measured to be red. If we choose to measure the second q-on’s color, we will surely get red. But as we discussed earlier, when introducing complementarity, if we choose to measure a q-on’s shape, when it is in the “red” state, we will have equal probability to find a square or a circle.

 

[PeterDonis]

According to the following paragraphs from the links I gave

Neither of which are valid references. You need to be looking at textbooks and peer-reviewed papers, not Wikipedia and pop science magazines.

 

[DrChinese]

Suppose Alice measures the z-spin and obtains +z, so that the quantum state collapses into state I. Now, instead of measuring the z-spin as well, Bob measures the x-spin. According to quantum mechanics, when the system is in state I, Bob's x-spin measurement will have a 50% probability of producing +x and a 50% probability of -x.

Of course, this is correct as written. If Alice measures the z spin and gets +z (a 50-50 likelihood), Bob can measure on the x, y or z planes and his will result will ALSO be completely random (regardless of which plane he chooses). If he chooses z to measure, his result will be random but correlated (as expected) with Alice's result.

 

[PeroK]

It seems as if information has propagated (faster than light) from Alice's apparatus to make Bob's positron assume a definite spin in the appropriate axis.

That's nonsense. Although, I wonder whether the article you are quoting goes on to say as much, and is playing devil's advocate at this point.

Quanta Magazine link: -

Upon deeper reflection, the paradox dissolves further. Indeed, let us consider again the state of the second system, given that the first has been measured to be red. If we choose to measure the second q-on’s color, we will surely get red. But as we discussed earlier, when introducing complementarity, if we choose to measure a q-on’s shape, when it is in the “red” state, we will have equal probability to find a square or a circle.

Again, q will get square or circle with 50% probability regardless of what p does or does not do.

 

[LozD]

This is not an entangled state. In an entangled state, the particles' entangled properties do not have a specific value until they are measured. So you would actually have something more like:

Before measurement:
50% (A=1 + B=0) + 50% (A=0 + B=1)

After measurement:
100% (A=1, B=0) [or vice versa 100% (A=0, B=1)]

Note that this assumes you are measuring the 2 particles in the same manner. The order of measurement is not important to the statistical results.

Aha, now you're talkin'!

OK, my supposition that the probabilities could be set across two properties of the state in preparing the entanglement might be an issue. I think this more technical / mathematical discussion might be the same thing. From skim reading it, it appears the "correct" answerer is saying the preparing can set the probabilities but you lose something else in the maths or something that spoils it. Let's read it a bit more slowly ...

https://physics.stackexchange.com/q...pare-and-reset-probabilities-to-send-informat

 

[Irishdoug]

Many entangled particle pairs are prepared with property A (1 or 0) at 99% probability of 1, and property B (1 or 0) at 1% probability of 1.

I don't believe such a state would really be considered entangled.

 

[Nugatory]

Many entangled particle pairs are prepared with property A (1 or 0) at 99% probability of 1, and property B (1 or 0) at 1% probability of 1.

I don't believe such a state would really be considered entangled.

The state $\sqrt{.99}|11\rangle+\sqrt{.01}|00\rangle$ would, I think, generally be considered to be entangled. (I also have no idea how to prepare it).

 

[LozD]

To be honest, the B initial B state doesn't matter. That was an unnecessary constraint I applied. B can just be 50% with no preparing. All that's needed is for A to be able to be shaped to something that isn't 50%. Does that make for an entanglement now then, with only A "nudged"?

 

[Nugatory]

To be honest, the B initial B state doesn't matter.

On the contrary, what you are calling “the B state” is essential to calculating the results of B’s measurement. It is also an essential part of the overall system state because, as you said in #4 above, the two particles share the same state: there is a single quantum state of the entire two-particle system, it includes all the information needed to calculate the result of measurements by A and B, and because it cannot be separated into an “A state” and a “B state” (formally this property is called “non-factorizable” and it what makes the state entangled) it makes no sense to talk about either particle’s state in isolation.

 

[LozD]

On the contrary, what you are calling “the B state” is essential to calculating the results of B’s measurement. It is also an essential part of the overall system state because, as you said in #4 above, the two particles share the same state: there is a single quantum state of the entire two-particle system, it includes all the information needed to calculate the result of measurements by A and B, and because it cannot be separated into an “A state” and a “B state” (formally this property is called “non-factorizable” and it what makes the state entangled) it makes no sense to talk about either particle’s state in isolation.

But we don't care about the results of B's measurement. We only care that its has been measured or not.

Whenever I refer to measuring at one end or the other I am assuming that that action is measuring the shared state. There is no local state if they are entangled, I get that, I misled you if I made you think I was referring to either particle in isolation.

I think my misunderstanding might be to do around state and what I'm calling "properties". The state is the "wholeness" of the shared characteristics of the pair; the state has sub-components like axis, velocity, location, etc. All those properties make up the state. Or am I wrong there?

Hence why the Quanta Magazine article refers to the two systems (the two particles in my parlance) each having shape and colour as two properties that make up the state. And the Wiki refers to x and y-axis spin making up the shared state of the pair in their example.

 

[PeroK]

I think my misunderstanding might be to do around state and what I'm calling "properties". The state is the "wholeness" of the shared characteristics of the pair; the state has sub-components like axis, velocity, location, etc. All those properties make up the state. Or am I wrong there?

Look at it this way: I have a particle and I am about to measure some quantity associated with that particle. Let's say spin about the x-axis to be precise. And, let's say, that the probabililty is 50-50 that it is spin-up or spin-down.

Now, you tell me that something about another particle has been measured. That cannot change the 50-50. You might say: someone measured the x-spin of another particle. Okay, but it's still 50-50 on my particle.

Then, however, you tell me the result of that measurement. You say: the other particle was measured and it was spin-up. That changes the probability. Now I know that my particle is x spin-up.

Your mistake is to think that the probabilities change by knowing that another measurement has been made, without knowing the outcome.

That's fundamentally wrong, whatever you think you've read in Quanta magazine.

 

[LozD]

Look at it this way: I have a particle and I am about to measure some quantity associated with that particle. Let's say spin about the x-axis to be precise. And, let's say, that the probabililty is 50-50 that it is spin-up or spin-down.

Now, you tell me that something about another particle has been measured. That cannot change the 50-50. You might say: someone measured the x-spin of another particle. Okay, but it's still 50-50 on my particle.

Then, however, you tell me the result of that measurement. You say: the other particle was measured and it was spin-up. That changes the probability. Now I know that my particle is x spin-up.

Your mistake is to think that the probabilities change by knowing that another measurement has been made, without knowing the outcome.

That's fundamentally wrong, whatever you think you've read in Quanta magazine.

Your mistake is to think that the probabilities change by knowing that another measurement has been made, without knowing the outcome.

I'm not saying the probabilities change by knowing that another measurement has changed. I'm saying that the probability on, say, the x-axis spin change when the y-axis spin is measured, whether or not anyone knows about that measuring happening or not.

 

[Nugatory]

I think my misunderstanding might be to do around state and what I'm calling "properties".

There’s no way of responding without at least sniffing around the underlying math. If you’re not up for taking on an intro textbook you might give Giancarlo Girardi’s book “Sneaking a look at god’s cards” a try.

However I can try a very abbreviated corner-cutting explanation. In post #16 I wrote down an entangled state $|\psi\rangle$ that seems close to what you were trying to describe in your first post: $|\psi\rangle=\sqrt{.99}|11\rangle+\sqrt{.01}|00\rangle$

The things in $|\rangle$ brackets are called ”kets”; they are abstract mathematical objects that work like vectors - the sum of two kets is a ket, and a ket can be multiplied by a number to yield a “bigger” or “smaller” ket that “points in the same direction”.

The ket $|11\rangle$ corresponds to the state “If A measures the spin on a specific chosen in advance axis there is a 100% probability that they will get a 1 and if B measures the state on that same axis there is a 100% probability that they will get a 1”. The ket $|00\rangle$ is the same thing for 0 results.

The ket $|\psi\rangle$ is the state of the system and it is a sum of two kets, called a superposition. If we measure the spin of either particle on the specified axis the wave function will collapse to either $|00\rangle$ or $|11\rangle$; the probabilities are given by the square of the coefficients, in this case 99% and 1%.

So let’s say that A performs a measurement. 99% of the time the wave function collapses to $|11\rangle$ and 1% of the time to $|00\rangle$. Thus, after A has performed their measurement if B measures any property of their particle they will get results consistent with their particle being in the 1 state 99% of the time and consistent with their particle being in the 0 state 1% of the time. (And note that that is “any property” - if they measure something other than spin on the designated axis we’ll need to do some math to calculate the probabilities of those results from the 1 state and the 0 state, but we know how to do that).

But suppose A doesn’t perform any measurement so the wave function is uncollapsed when it gets to B? B does something to their particle, that collapses the wave function with the same 99% $|11\rangle$ and 1% $|00\rangle$ probabilities, and we are in the exact same situation as if A had made a measurement.

And that is why nothing A does can affect anything B observes, and therefore why A cannot send a signal to B by manipulating their side of the entanglement.

 

[LozD]

Aha, now you're talkin'!

OK, my supposition that the probabilities could be set across two properties of the state in preparing the entanglement might be an issue. I think this more technical / mathematical discussion might be the same thing. From skim reading it, it appears the "correct" answerer is saying the preparing can set the probabilities but you lose something else in the maths or something that spoils it. Let's read it a bit more slowly ...

https://physics.stackexchange.com/q...pare-and-reset-probabilities-to-send-informat

OK, the maths in that link I don't fully get, but the gist of the text description around it I have some inkling I think.

The answerer's answer to 1. acknowledges that you can "shape" (my terminology, sorry!) the probabilities on the particles and retain the entanglement. So 75/25% instead of 50/50% and they are entangled.

In questioner's 2 and 3 I can see now he is trying to "transmit" on just one property, like my property B, say x-axis spin or whatever. He's simply saying, if he shapes B to 75/25% and the remote observer measures enough of the particles and finds they're in the ration 75/25 then interpret a 1, else if he shapes B to 50/50% then the observer can interpret a 0. That simpler than my dual property approach and I never even thought of it.

The answerer demolishes it through the maths I don't understand, but I think basically says "No, the probability reverts to 50/50% no matter what you shape because the diagonal filter is a measurement and that collapses the entanglement.

But interestingly, in 4 the answerer talks about the internal entanglement between different properties of the state. So entanglement is possible between A and B (x-spin and y-spin) within the state, not just A (x-spin) to A (x-spin) and / or B (y-spin) to B (y-spin) between two particles that share a state.

Here's the paragraph ...

4. Each independent degree of freedom can sustain correlations with other degrees of freedom, on the same particle or other particles, in principle. In this case, all of the entanglement that the system had, aside from correlated uncertainties in their momenta (which you implicitly perform a weak measurement on, just by virtue of making a polarization measurement at a particular location after it was emitted), are in the polarizations. As the polarization of each beam is entirely involved in what entanglement there is (in a multi-particle scenario, you can consider entanglement spread across the system in a way that some particles are more involved in the entangled state than others), any single measurement will destroy all entanglement present.

"Any measurement will destroy all entanglement present". That's exactly what I am saying, is it not? That means if property A was shaped 99/1% (1/0) as I proposed, the measurement of B by the remote particle observer Y collapses the whole entanglement shooting match. So A becomes 50/50%. And that signifies a 1 was sent by Y (she measured B), whereas if A remains 99/1% that signifies a 1 was sent by Y (she didn't measure B).

 

[LozD]

When the entanglement collapses, I wonder what the speed of action is between the measured property that triggers the collapse and the other different properties with the same state (answerer's 4 above) to reset them - like if x-spin is measured how long does it take for y-spin to register that and go back to 50/50% if it was 99/1%? We know it's instantaneous for a property (say x-spin) between the two particles in the entangled state. But is it instantaneous between the measured property and the other properties in the state when collapse occurs?

Let's say my theory is true - the non-measured properties revert to 50/50% (1,0) from 99/1% (1,0) or whatever they were when entangled. Then the inter-property disentanglement transmission speed can't be instantaneous because then special relativity would be contradicted and causality would be transmittable faster than the speed of light. But if the inter-property disentanglement transmission speed was slower such that it exactly compensated for the instantaneousness of the inter-particle transmission then special relativity and the casualty faster than the speed of light problem would disappear.

So if the particles were separated by a distance such that the speed of light would take 1 minute to get between them, then even though the inter-particle collapse is instant the inter-property collapse would take 1 minute. So any event (the legendary cat with its poison trigger for example) hanging off the back of the switch back to 50/50% from 99/1% would still not be triggered before light would reach it from the remote particle and observer, because the delay has been shifted to the inter-property transmission.

That's testable, but not easy!

So while we wouldn't have a system that transmits information faster than the speed of light, we do have one that doesn't use light, electromagnetics, etc. mechanisms between its end points. So no-one could spy on it. Without access to one of the particles in the entanglement the communication would be secure (unless someone discovers what the transmission mechanism is for disentanglement). That'd be worth a bob or two if someone could get it working.

 

[StevieTNZ]

When the entanglement collapses, I wonder what the speed of action is between the measured property that triggers the collapse and the other different properties with the same state (answerer's 4 above) to reset them - like if x-spin is measured how long does it take for y-spin to register that and go back to 50/50% if it was 99/1%? We know it's instantaneous for a property (say x-spin) between the two particles in the entangled state. But is it instantaneous between the measured property and the other properties in the state when collapse occurs?

Let's say my theory is true - the non-measured properties revert to 50/50% (1,0) from 99/1% (1,0) or whatever they were when entangled.

I think you really need to read "Sneaking a Look at God's Cards" as suggested in post #22. Or https://mitpress.mit.edu/books/quantum-entanglement

 

[LozD]

I think you really need to read "Sneaking a Look at God's Cards" as suggested in post #22. Or https://mitpress.mit.edu/books/quantum-entanglement

For the shear brilliance of the name "Sneaking a Look at God's Cards" my order is on its way!

 

[Nugatory]

I'm saying that the probability on, say, the x-axis spin change when the y-axis spin is measured, whether or not anyone knows about that measuring happening or not.

They don’t, if we’re measuring x on one particle and y on the other.

The probability of getting spin-up on x given that a y measurement has been made on the other particle is the sum of the probability of getting spin-up on x if y was measured up times the probability that y would be measured spin up and the probability of spin-up on x if y was measured down times the probability of y would be measured spin-down - and that is equal to the probability of getting spin-up on x if no measurement has been made on y.

 

[DrChinese]

1. When the entanglement collapses, I wonder what the speed of action is between the measured property that triggers the collapse... That's testable, but not easy!

2. Without access to one of the particles in the entanglement the communication would be secure (unless someone discovers what the transmission mechanism is for disentanglement). That'd be worth a bob or two if someone could get it working.

The mystery here is why you are speculating about nuances of entanglement when you don't know the basics. Many of the questions you ask would be answered if you knew more about the subject from a scientific perspective (rather than reading highly abbreviated summaries). You are obviously interested and intelligent, so you should take time to follow some of the paths that have already been tread by others. Some basics are in the following paper, which is intended for college undergrads. It in turn references 3 key papers (and more of course): its references 4 (Einstein et al, 1935), 25 (Bell, 1964) and 30 (Aspect et al, 1982) - are considered essential to any sort of understanding of entanglement. And yes, it would take some of your time to learn about these and why they are important. But considering that there have been over 30,000 papers written on entanglement in just the past 20 years since the one below was written in 2002, reading these I mention will be a good jump start.

https://arxiv.org/abs/quant-ph/0205171
"We use polarization-entangled photon pairs to demonstrate quantum nonlocality in an experiment suitable for advanced undergraduates. The photons are produced by spontaneous parametric down conversion using a violet diode laser and two nonlinear crystals. The polarization state of the photons is tunable. Using an entangled state analogous to that described in the Einstein-Podolsky-Rosen “paradox,” we demonstrate strong polarization correlations of the entangled photons. Bell’s idea of a hidden variable theory is presented by way of an example and compared to the quantum prediction. A test of the Clauser, Horne, Shimony and Holt version of the Bell inequality finds S = 2.307±0.035, in clear contradiction of hidden variable theories. The experiments described can be performed in an afternoon."

----------------------------------------

Further, as you might imagine, many facets of entanglement have been studied quite deeply - including many around what you are speculating (although these studies relate to known theory and experiment).

1. Although not exactly the same as your 1 (per my labeling): the minimum experimentally determined speed at which collapse occurs between entangled particles is 10,000 times the speed of light (10^4 c, or 4 orders of magnitude). It is probably instantaneous, but we cannot be absolutely certain. :)

https://arxiv.org/abs/0808.3316

2. And secure communication is in fact one of the objectives in quantum cryptography:

https://en.wikipedia.org/wiki/Quantum_cryptography

Good luck!

 

[PeroK]

Your mistake is to think that the probabilities change by knowing that another measurement has been made, without knowing the outcome.

I'm not saying the probabilities change by knowing that another measurement has changed. I'm saying that the probability on, say, the x-axis spin change when the y-axis spin is measured, whether or not anyone knows about that measuring happening or not.

Okay, I think I understand your model of entanglement. We have entangled particles A and B:

Particle A is definitely x-spin-up (or 99% if you prefer); particle B is definitely/99% x-spin-down.

You measure z-spin of particle A. Now particle A is 50% x-spin-up; and, particle B is now 50% x-spin-up.

Therefore: if you measure particle B before particle A has been measured you almost certainly get x-spin-down; but, if you wait until particle A has been measured (in the z-direction), then the state of particle B changes from 99% x-spin-down to 50% spin-down.

This allows someone to send a message FTL by choosing to measure the z-spin of particle A or not.

I think that's what you're saying.

The only problem with this is that this is not QM; this is LozD mechanics (LM), which allows FTL signalling. There is very little, if anything, in common between QM and LM. They are two entirely different physical theories! And, critically, the experimental evidence supports QM and not LM.

 

[LozD]

Okay, I think I understand your model of entanglement. We have entangled particles A and B:

Particle A is definitely x-spin-up (or 99% if you prefer); particle B is definitely/99% x-spin-down.

You measure z-spin of particle A. Now particle A is 50% x-spin-up; and, particle B is now 50% x-spin-up.

Therefore: if you measure particle B before particle A has been measured you almost certainly get x-spin-down; but, if you wait until particle A has been measured (in the z-direction), then the state of particle B changes from 99% x-spin-down to 50% spin-down.

This allows someone to send a message FTL by choosing to measure the z-spin of particle A or not.

I think that's what you're saying.

The only problem with this is that this is not QM; this is LozD mechanics (LM), which allows FTL signalling. There is very little, if anything, in common between QM and LM. They are two entirely different physical theories! And, critically, the experimental evidence supports QM and not LM.

LM rocks!

Yes, that's it exactly! My poor usage of terminology confused everyone.

So my question is: why is LM not QM? What am I doing wrong in the LM? Now that I've cleared up my confused attempt to pose a thought experiment and been understood, what specifically is actually wrong in the LM trail of logic?